Direct image of double cover. Let $\mathbb{Z}$ be the constant sheaf on $\mathbb{S}^1$, $f: \mathbb{S}^1 \to\mathbb{R}P^1$ the double cover and $\mathcal{A} = f\mathbb{Z}$. 
Then $\mathcal{A}$ has stalks $\mathbb{Z}\oplus\mathbb{Z}$ but I am having difficulties describing the topology of $\mathcal{A}$. That is, I do not understand why the sheaf is 'twisted' via the automorphism $(n,k)\to (k,n)$. 
edit: my idea was flawed so I might as well remove it.
 A: Locally, $\mathcal A$ is a constant sheaf with stalks $\mathbb Z \oplus\mathbb Z$.
More precisely, if $U$is an open of $S^1$, $\underline {\mathbb Z}(U)$ is the product of one copy of $\mathbb Z$ for each connected component of $U$, and if $V$ is an open of $\mathbb P^1$, then $\mathcal A (V) = \underline{\mathbb Z}(f^{-1} (V))$ = the product of as many copies of $\mathbb Z$ as there are connected components of $f^{-1} (V)$, which is usually twice the number of connected components of $V$.
For example, if we take two distinct points $x_1,x_2 \in \mathbb P^1$, and take $V_i = \mathbb P^1 - \{x_i\}$, then $\mathcal A |_{V_i}$ is a constant sheaf with stalks $\mathbb Z \times\mathbb Z$. (the connected components of $f^{-1}(U)$ are always twice the connected components of $U$ when $U$ is an open subset of $V_i$)
But, as a whole, $\mathcal A$ is not the constant sheaf on $\mathbb P^1$, because $\mathcal A(\mathbb P^1) = \underline {\mathbb Z} (S^1) = \mathbb{Z}$, which is not $\mathbb Z \oplus \mathbb Z$.
We can describe it instead as a locally constant sheaf, which is constant on $V_1$ and $V_2$, and we need to describe how it glues.
Let $W = V_1 \cap V_2$.$W$ has two components, call them $W = W_a \cup W_b$.  The fact that $\mathcal A$ is constant on $V_i$ means that it gives two different isomorphisms from $\mathcal A(W_a) \to \mathcal A(W_b)$, one going through $V_1$ and the other going through $V_2$.
Denote $W_a^1, W_a^2$ the connected components of $f^{-1}(W_a)$, do the same for $W_b^1, W_b^2$, and put the exponents such that $W_a^1$ and $W_b^1$ are in the same connected component of $f^{-1}(V_1)$, so that the connected components of $f^{-1}(V_1)$ correspond to $W_a^1 \cup W_b^1$ and $W_a^2 \cup W_b^2$, and those of $f^{-1}(V_2)$ correspond to $W_a^1 \cup W_b^2$ and $W_a^2 \cup W_b^1$
Then, we have restriction isomorphisms $\rho_i^a : \mathcal A(V_i) \to \mathcal A(W_a)$, which look like this : $\rho_1^a(x,y) = (x,y)$ and $\rho_2^a(x,y) = (x,y)$ ; and $\rho_i^b : \mathcal A(V_i) \to \mathcal A(W_b)$ : $\rho_1^a(x,y) = (x,y)$ and $\rho_2^a(x,y) = (y,x)$.
Then, the twisting map is the isomorphism $\tau = \rho_2^a \circ (\rho_2^b)^{-1} \circ \rho_1^b \circ (\rho_1^a)^{-1} : \mathcal A(W_a) \to \mathcal A(W_a) $. You should get that $\tau(x,y) = (y,x)$.
This isomorphism describes what happens to sections on $W_a$ when you go once through $\mathbb P^1$ : the two connected components of $f^{-1}(W_a)$ get switched around, and you are basically right when you describe it in your comment.
And in fact, this map is enough to recover all the information you need to describe the sheaf $\mathcal A$ completely, so we can describe $\mathcal A$ as a locally constant sheaf whose stalks are $\mathbb Z \oplus \mathbb Z$ twisted by $\tau$ when we do one loop around $\mathbb P ^1$
A: Let's recall some general facts.
Fact 1. If $\mathscr{F}$ is a sheaf on $X$ and $f : X \to Y$ is a continuous map, then the direct image sheaf  $f_* \mathscr{F}$ is the presheaf defined by $U \mapsto \mathscr{F}(f^{-1} U)$. In other words, a section of $f_* \mathscr{F}$ over $U$ is the same thing as a section of $\mathscr{F}$ over $f^{-1} U$.
Fact 2. If $\mathscr{F}$ is a sheaf on $X$, then its éspace étalé $E$ is constructed as follows: as a set,
$$E = \coprod_{x \in X} \mathscr{F}_x$$
where $\mathscr{F}_x$ denotes the stalk of $\mathscr{F}$ at $x$, and the topology of $E$ is the one generated by the base
$$\hat{s} = \{ (x, s_x) : x \in \operatorname{dom}(s) \}$$
for all local sections $s$ of $\mathscr{F}$. Consequently, it is possible to express $E$ as a colimit of open subsets of $X$.

Now, if $X$ is Hausdorff and $f : X \to Y$ is a finite covering map, then it is very easy to describe the stalk of $f_* \mathscr{F}$ at a point $y$: it is simply the product of the stalks $\mathscr{F}_x$ as $x$ varies over $f^{-1} \{ y \}$. (If $f$ does not have finite discrete fibres, something more complicated happens.) The way to think about this is that the germ of a section of $f_* \mathscr{F}$ at $y$ must specify the germs of a section of $\mathscr{F}$ at all the preimages of $y$.
Now, let $g : X \to X$ be an automorphism over $Y$. It is not quite true that this induces an automorphism of every sheaf on $X$. But for constant sheaves it does: if $E$ is the éspace étalé of a constant sheaf $\mathscr{F}$ and $E = X \times F$ for some discrete space $F$, then the induced automorphism on $E$ is the obvious one, i.e. $g \times F : X \times F \to X \times F$. Therefore there is an induced automorphism on the stalks of $f_* \mathscr{F}$, which just permutes the factors of the product $\prod_{x \in f^{-1} \{ y \}} \mathscr{F}_x$. (More generally, this holds when $\mathscr{F} = f^* \mathscr{G}$ for some sheaf $\mathscr{G}$ on $Y$ – then $\mathscr{F}$ inherits an action by the automorphisms of $X$.)
